Related lecture material This derivation is part of Lecture 2: Viscosity and Momentum Diffusion Physical Setup A semi-infinite fluid initially at rest. At , a plate at suddenly moves with velocity . Initial and Boundary Conditions: Misplaced &\vec{u}(x,y,t=0) &= 0 \\ \text{At } t = 0^{+}, \quad u(y=0) &= u_0 \end{align}$$ --- ## Simplifying Assumptions We assume spatial uniformity in the $x$-direction: $$\boxed{\frac{\partial()}{\partial x} = 0}$$ Ignoring any initial transients, this leads to: From continuity equation: $$\frac{\partial v}{\partial y} = 0$$ Combined with no-slip at the plate: $$v(y=0) = 0 \quad \Rightarrow \quad v = 0 \quad \text{for all } y$$ --- ## Governing Equations $x$-momentum equation: $$\rho \frac{\partial u}{\partial t} = - \frac{\partial p}{\partial x} + \mu \frac{\partial^2 u}{\partial y^2}$$ $y$-momentum equation: $$\frac{\partial p}{\partial y} = 0$$ Pressure gradient: For an open container with $(p,u,v) = 0$ far from the plate: $$\frac{\partial p}{\partial x} = 0$$ Simplified momentum equation: $$\boxed{\rho \frac{\partial u}{\partial t} = \mu \frac{\partial^2 u}{\partial y^2}}$$ or equivalently: $$\boxed{\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}}$$ where $\nu = \mu/\rho$ is the kinematic viscosity. > [!note] Key Insight > > We cannot directly integrate this PDE as $u = f(y,t)$. We need a more sophisticated approach. --- ## Boundary Conditions $$\begin{align} u(y,0) &= 0 \quad \text{(initially at rest)} \\ u(0,t) &= u_0 \quad \text{(plate velocity)} \\ u(\infty,t) &= 0 \quad \text{(far-field BC)} \end{align}$$ --- ## Similarity Solution Approach Using dimensional analysis, we seek a solution of the form: $$u = \phi\left(u_0, y, t, \frac{\mu}{\rho} = \nu\right)$$ By dimensional reasoning, the velocity profile should depend on the similarity variable: $$\frac{u}{u_0} = \phi\left(\frac{y}{\sqrt{\nu t}}\right)$$ > [!important] Similarity Variable > > The characteristic length scale grows as $\sqrt{\nu t}$, reflecting the diffusive nature of momentum transport. Define the similarity variable: $$\boxed{\eta = \frac{y}{2\sqrt{\nu t}}}$$ and the dimensionless velocity: $$\boxed{\frac{u}{u_0} = F(\eta)}$$ --- ## Transformation to ODE We now transform the PDE by computing the necessary derivatives. ### Time derivative: $$\frac{\partial u}{\partial t} = u_0 \frac{dF}{d\eta} \frac{\partial \eta}{\partial t}$$ where: $$\frac{\partial \eta}{\partial t} = -\frac{1}{4} \frac{y}{\sqrt{\nu} t^{3/2}} = -\frac{\eta}{2t}$$ Therefore: $$\frac{\partial u}{\partial t} = u_0 F' \cdot \left(-\frac{\eta}{2t}\right) = \frac{-u_0 F' \eta}{2t}$$ ### Spatial derivatives: $$\frac{\partial u}{\partial y} = u_0 \frac{dF}{d\eta} \frac{\partial \eta}{\partial y} = u_0 F' \frac{1}{2\sqrt{\nu t}}$$ $$\frac{\partial^2 u}{\partial y^2} = u_0 F'' \frac{1}{2\sqrt{\nu t}} \frac{\partial \eta}{\partial y} = \frac{u_0 F''}{4\nu t}$$ ### Substituting into the PDE: $$\frac{-u_0 F' \eta}{2t} = \nu \cdot \frac{u_0 F''}{4\nu t}$$ Simplifying: $$\boxed{F'' + 2\eta F' = 0}$$ This is an ordinary differential equation for $F(\eta)$. --- ## Transformed Boundary Conditions  The original boundary conditions become: $$\begin{align} \eta = \frac{y}{2\sqrt{\nu t}} &\quad \Rightarrow \quad \begin{cases} t = 0: & \eta \to \infty \\ y \to \infty: & \eta \to \infty \\ y = 0: & \eta = 0 \end{cases} \end{align}$$ Therefore: $$\begin{align} F(\eta \to \infty) &= 0 \\ F(\eta = 0) &= 1 \end{align}$$ --- ## Solving the ODE Starting with: $$F'' + 2\eta F' = 0$$ Rearrange: $$\frac{dF'}{d\eta} = -2\eta F'$$ Separate variables: $$\frac{dF'}{F'} = -2\eta \, d\eta$$ Integrate: $$\ln(F') = -\eta^2 + C$$ Exponentiate: $$F' = A e^{-\eta^2}$$ Integrate again: $$\boxed{F = A \int_0^{\eta} e^{-s^2} \, ds + B}$$ --- ## Applying Boundary Conditions At $\eta = 0$: $F(0) = 1$ $$B = 1$$ As $\eta \to \infty$: $F(\infty) = 0$ $$0 = A \int_0^{\infty} e^{-s^2} \, ds + 1$$ Using the Gaussian integral: $$\int_0^{\infty} e^{-s^2} \, ds = \frac{\sqrt{\pi}}{2}$$ We get: $$0 = A \frac{\sqrt{\pi}}{2} + 1 \quad \Rightarrow \quad A = -\frac{2}{\sqrt{\pi}}$$ --- ## Final Solution Substituting the constants: $$\boxed{F(\eta) = 1 - \frac{2}{\sqrt{\pi}} \int_0^{\eta} e^{-s^2} \, ds}$$ ### Error Function Representation The error function is defined as: $$\erf(\eta) = \frac{2}{\sqrt{\pi}} \int_0^{\eta} e^{-s^2} \, ds$$ Therefore: $$\boxed{F(\eta) = 1 - \erf(\eta)}$$ or equivalently, using the complementary error function $\erfc(\eta) = 1 - \erf(\eta)$: $$\boxed{\frac{u}{u_0} = \erfc\left(\frac{y}{2\sqrt{\nu t}}\right) = 1 - \erf\left(\frac{y}{2\sqrt{\nu t}}\right)}$$ --- ## Physical Interpretation > [!summary] Key Results > > > 1. Diffusive spreading: The velocity profile spreads into the fluid with a characteristic length scale $\delta(t) \sim \sqrt{\nu t}$ > > 2. Self-similar evolution: At different times, the velocity profiles collapse onto a single curve when plotted against $\eta = y/(2\sqrt{\nu t})$ > > 3. Momentum diffusion: The kinematic viscosity $\nu$ acts as a diffusivity for momentum, exactly analogous to thermal or mass diffusion > > 4. Penetration depth: At any given time, most of the velocity change occurs within $y \lesssim \sqrt{\nu t}$ The solution demonstrates that viscosity is fundamentally a diffusive process for momentum transport, validating the kinetic theory picture developed in [[2-What-is-Viscosity|Lecture 2]]. ![[StokesFirstProblem.pdf]]
